Write Up – Classic Cipher 2

Write Up – Classic Cipher 2

Challenge Overview

Category: Crypto
Difficulty: Easy
Author: [Unknown]

Diberikan sebuah file clasic2.txt yang berisi ciphertext sangat panjang (2815 bytes). Challenge ini melibatkan multiple encoding layers yang lebih kompleks dari Classic 1, menggunakan kombinasi Base58, Base62, dan binary mapping.

Format flag:

FGTE{...}

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Files Given

1. Ciphertext (clasic2.txt)

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
(2815 bytes total)

String ini tidak terlihat seperti Base64 standar karena mengandung karakter lowercase dan uppercase yang bercampur.

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Key Insight

Layered Encoding dengan Base58 dan Base62

Ciphertext menggunakan multiple encoding layers yang tidak umum:

1. Base62 (layer pertama) - encoding menggunakan alfabet 0-9A-Za-z
2. Base58 (layer kedua) - encoding Bitcoin-style tanpa karakter ambigu (0, O, I, l)
3. Binary mapping dengan nilai dominan: 88 → 0 dan 99 → 1

Analisis:

• String mengandung karakter alphanumeric (Base62 charset)
• Setelah decode Base62 → menghasilkan data yang cocok dengan Base58 charset
• Decode Base58 → menghasilkan deretan angka
• Angka-angka tersebut memiliki 2 nilai dominan: 88 dan 99
• Mapping binary: 88→0, 99→1 untuk mendapatkan bitstring → ASCII

Kesimpulan: Harus melakukan 3-stage decoding: Base62 → Base58 → Binary mapping (88/99) → ASCII

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Solution Steps

Step 1 – Base62 Decode

Decode string Base62 menggunakan alfabet 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz:

import base64

B62_ALPH = b"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"

def base62_decode(s: bytes):
    idx = {c: i for i, c in enumerate(B62_ALPH)}
    n = 0
    for c in s:
        n = n * 62 + idx[c]
    return n.to_bytes((n.bit_length() + 7) // 8 or 1, "big")

with open("clasic2.txt", "rb") as f:
    data = f.read().strip()

stage1 = base62_decode(data)

Step 2 – Base58 Decode

Decode hasil Base62 dengan Base58 (Bitcoin alphabet):

B58_ALPH = b"123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"

def base58_decode(s: bytes):
    idx = {c: i for i, c in enumerate(B58_ALPH)}
    n = 0
    for c in s:
        n = n * 58 + idx[c]
    # Handle leading '1' sebagai leading zero bytes
    lead = sum(1 for c in s if c == ord('1') else 0)
    out = n.to_bytes((n.bit_length() + 7) // 8 or 1, "big")
    return b"\x00" * lead + out

stage2 = base58_decode(stage1)

Hasil: Deretan angka (dalam bytes) yang didominasi nilai 88 dan 99

Step 3 – Binary Mapping

Extract angka-angka dan mapping ke bits:

• Nilai 88 → 0
• Nilai 99 → 1

# Filter hanya byte 88 dan 99
bits = "".join("0" if b == 88 else "1" if b == 99 else "" for b in stage2)

Step 4 – Bits to Bytes

Convert bitstring ke bytes, kemudian decode sebagai ASCII:

# Pastikan panjang bits kelipatan 8
bits = bits[: (len(bits) // 8) * 8]

# Convert setiap 8 bits menjadi 1 byte
out = bytes(int(bits[i:i+8], 2) for i in range(0, len(bits), 8))

print(out.decode())

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Solve Script

#!/usr/bin/env python3
from pathlib import Path

B62_ALPH = b"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
B58_ALPH = b"123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"

def baseN_decode(s: bytes, alph: bytes):
    idx = {c: i for i, c in enumerate(alph)}
    n = 0
    for c in s:
        n = n * len(alph) + idx[c]
    out = n.to_bytes((n.bit_length() + 7) // 8 or 1, "big")
    # Handle leading zeros for base58
    if alph == B58_ALPH:
        lead = sum(1 for c in s if c == ord('1') else 0)
        out = b"\x00" * lead + out
    return out

# Read ciphertext
data = Path("clasic2.txt").read_bytes().strip()

# Stage 1: Base62 decode
stage1 = baseN_decode(data, B62_ALPH)

# Stage 2: Base58 decode
stage2 = baseN_decode(stage1, B58_ALPH)

# Stage 3: Binary mapping (88->0, 99->1)
bits = "".join("0" if b == 88 else "1" if b == 99 else "" for b in stage2)

# Stage 4: Bits to bytes
bits = bits[: (len(bits) // 8) * 8]
out = bytes(int(bits[i:i+8], 2) for i in range(0, len(bits), 8))

print(out.decode())

Output:

FGTE{L4njut4n_Cl4sic_2_b4s3_58_d4n_b4s3_62}

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Conclusion

Challenge ini menggunakan layered encoding dengan kombinasi encoding yang jarang digunakan:

1. Base62 decoding (alfabet 0-9A-Za-z)
2. Base58 decoding (Bitcoin-style alphabet)
3. Custom binary mapping (88/99 → binary → ASCII)

Kunci dari challenge ini adalah mengenali Base62 dan Base58 encoding yang tidak umum dalam CTF, serta menemukan pola binary mapping dengan 2 nilai dominan (88 dan 99).

Pelajaran: Base58 dan Base62 sering digunakan dalam blockchain dan URL shortening. Dalam CTF, perhatikan distribusi byte - jika ada 2 nilai yang sangat dominan (>85%), kemungkinan itu adalah binary encoding.

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